In this paper it is first shown that if M is a 2-dimensional
surface, then $M^{(m)}$ is orientable if and only if M is orientable.
Using the Macdonald\u27s result [5].
the Betti numbers of $M^{(m)}$ are expressed explicitely,
the Euler characteristic is found and it depends only on $m$
and the Euler characteristic of M.
Moreover, the formula for the Euler characteristic is proved alternatively also without using the Macdonalds\u27s results

Kostadin Trenčevski

English

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MATHEMATICAL COMMUNICATIONS 295Math. Commun., Vol. 15, No. 2, pp. 295-301 (2010)The Euler characteristic of the symmetric product of a finiteCW-complex∗Kostadin Trenčevski1,†1 Institute of Mathematics, Ss. Cyril and Methodius University, P.O. Box 162, 1 000Skopje, MacedoniaReceived January 14, 2009; accepted September 3, 2009Abstract. In this paper it is first shown that if M is a 2-dimensional surface, then M (m) isorientable if and only if M is orientable. Using the Macdonald’s result [5] the Betti numbersof M (m) are expressed explicitely, the Euler characteristic is found and it depends only onm and the Euler characteristic of M . Moreover, the formula for the Euler characteristic isproved alternatively also without using the Macdonalds’s results.AMS subject classifications: 57N65Key words: symmetric products of manifolds, homology, Euler characteristic, orientability1. Some preliminaries about symmetric products of manifoldsLet M be an arbitrary set and m a positive integer. In the m-fold Cartesian productMm we define a relation ≈ such that(x1, · · · , xm) ≈ (y1, · · · , ym) ⇔there exists a permutation θ : {1, 2, · · · ,m} → {1, 2, · · · ,m} such that yi = xθ(i),(1 ≤ i ≤ m).This is a relation of equivalence and the class represented by (x1, · · · , xm) willbe denoted by [x1, · · · , xm] and the set Mm/ ≈ will be denoted by M (m). Theset M (m) is called a symmetric product of M . Some authors call it a permutationproduct of M .If M is a topological space, then M (m) is also a topological space. The spaceM (m) was introduced quite early [9], and it was studied in [8, 4, 7, 1, 10], and in thePh.D. thesis of Wagner [17]. Some recent results are obtained in [16, 2, 3, 11, 12].If M is an arbitrary connected manifold and m > 1, then it is proved in [9] thatπ1(M (m)) ∼= H1(M,Z). (1)Another important result [17] states that (Rn)(m) is a manifold only for n = 2. Ifn = 2, then (R2)(m) = C(m) is homeomorphic to Cm. Indeed, using that C is analgebraically closed field, it is obvious that the map ϕ : C(m) → Cm defined byϕ[z1, · · · , zm] = (σ1(z1, · · · , zm), σ2(z1, · · · , zm), · · · , σm(z1, · · · , zm))∗This paper is supported by the Ministry of Education and Science of the Republic of Macedonia.†Corresponding author. Email address: kostatre@pmf.ukim.mk (K. Trenčevski)http://www.mathos.hr/mc c©2010 Department of Mathematics, University of Osijek296 K.Trenčevskiis a bijection, where σi (1 ≤ i ≤ m) is the i-th symmetric function of z1, · · · , zm, i.e.σi(z1, · · · , zm) =∑1≤j1<j2<...<ji≤mzj1 · zj2 · · · zji .The map ϕ is also a homeomorphism. If M is a 1-dimensional complex manifold,then M (m) is a complex manifold, too. For example, if M is a 2-sphere, i.e. acomplex manifold CP 1, then M (m) is the projective complex space CPm. Using thesymmetric products it is easy to see how M (m) = CPm decomposes into disjointcells C0,C1, · · · ,Cm. Let ξ ∈ M . Then we define [x1, · · · , xm] ∈ Mi if exactly i ofthe elements x1, · · · , xm are equal to ξ. ThusM (m) = M0 ∪M1 ∪ · · · ∪Mm = (M \ ξ)(m) ∪ (M \ ξ)(m−1) ∪ · · · ∪ (M \ ξ)(0)= C(m) ∪ C(m−1) ∪ · · · ∪ C(0) = Cm ∪ Cm−1 ∪ · · · ∪ C0.This theory about symmetric products has an important role in the theory of topo-logical commutative vector valued groups [14, 15, 13].The Poincaré polynomial of a symmetric product of a compact polyhedron isgiven in [5]. If B0, B1, B2, · · · are the Betti numbers of a space M , then thePoincaré polynomial of the mth symmetric product of M is the coefficient in frontof tm in the power series expansion of(1 + xt)B1(1 + x3t)B3 · · ·(1− t)B0(1− x2t)B2(1− x4t)B4 · · · . (2)Moreover, the homology of the symmetric products has been determined completelyby J.Milgram [6].In this paper some consequences of the results of Macdonald are given and analternative proof for the expression of the Euler characteristic of the symmetricproduct of M is given, too.2. Some conclusions about the symmetric productsFirst, we prove the following theorem which gives a necessary and sufficient conditionfor the orientability of a 2-dimensional surface.Theorem 1. Let M be a 2-dimensional manifold. The manifold M (m) is orientableif and only if M is orientable.Proof. If M is orientable, then M is a complex manifold. Thus, M (m) is also acomplex manifold and hence M (m) is orientable.Assume that M is non-orientable. Then there exist a non-orientable open subsetT of M and an open subset U which is homeomorphic to R2, such that T ∩ U = ∅.Then M (m) contains the non-orientable submanifold T ×U (m−1) ∼= T ×R2m−2 withthe same dimension and hence M (m) is not orientable.For example, it is known that (RP 2)(m) = RP 2m, and both spaces RP 2 andRP 2m are not orientable.The Euler characteristic of the symmetric product 297Let M be a finite CW complex and let us denote byBi = dim(Hi(M,Z)), i = 0, 1, 2, · · · , n = dimMand let us denote the Betti numbers of the CW complex M (m) byB(m)i = dim(Hi(M(m),Z)), i = 0, 1, 2, · · · , nm.As a direct consequence of the result of Macdonald [5] we have the following twotheorems.Theorem 2. The Betti numbers B(m)i i = 0, 1, 2, · · · , nm, are given explicitly byB(m)i = (−1)i∑α0,··· ,αn(α0 − 1 + B0α0)·(α1 − 1−B1α1)×(α2 − 1 + B2α2)· · ·(αn − 1 + (−1)nBnαn), (3)where the summation over α0, · · · , αn is under the following restrictions0 ≤ α0, · · · , αn ≤ m, α0 + · · ·+ αn = m, α1 + 2α2 + · · ·+ nαn = i. (4)Proof. Using the equality(α− 1 + Bα)= (−1)α(−Bα),for a positive integer α and integer B, we should prove thatB(m)i = (−1)i+m∑α0,··· ,αn(−B0α0)·(B1α1)·(−B2α2)· · ·((−1)n+1Bnαn),where the summation over α1, · · · , αn is given by (4). Since(1 + xt)B1(1 + x3t)B3 · · ·(1− t)B0(1− x2t)B2(1− x4t)B4 · · · =∞∑m=0mn∑i=0B(m)i xitm,it is sufficient to prove that(1− t)−B0(1 + xt)B1(1− x2t)−B2(1 + x3t)B3(1− x4t)−B4 · · ·=∞∑m=0mn∑i=0∑α0,··· ,αn(−1)i+mxitm∑α0,··· ,αn(−B0α0)·(B1α1)×(−B2α2)· · ·((−1)n+1Bnαn),where the summation over α1, · · · , αn is given by (4). Notice that (−1)i+m= (−1)α0+α2+α4+···. Applying the binomial formula for the powers of the left-hand side and choosing the summand (−1)α0(−B0α0)tα0 from (1− t)−B0 , (B1α1)tα1xα1from (1 + xt)B1 , (−1)α2(−B2α2)tα2x2α2 from (1 − x2t)−B2 and so on, and sum overα0, α1, α2, · · · , according to (4), we obtain that the previous equality is true.298 K.TrenčevskiTheorem 3. Euler characteristics of M and the symmetric product M (m) are relatedbyχ(M (m)) =(m + χ(M)− 1m). (5)Proof.nm∑i=0(−1)iB(m)i =nm∑i=0∑α0,··· ,αn(−1)α0(−B0α0)(−1)α1(B1α1)· · · (−1)αn((−1)n+1Bnαn),where the summation over α0, · · · , αn satisfies conditions (4). Since α1 +2α2 + · · ·+nαn = i and we sum over i, using that α0 + · · ·+ αn = m we obtainχ(M (m)) =nm∑i=0(−1)iB(m)i =∑α0+···+αn=m(−1)m(−B0α0)(B1α1)· · ·((−1)n+1Bnαn)= (−1)m(−B0 + B1 −B2 + · · ·+ (−1)n+1Bnm)= (−1)m(−χ(M)m)=(m + χ(M)− 1m).Notice that Theorem 3 also follows from (2). Indeed if we put x = −1, then theEuler characteristic of M (m) is the coefficient in front of tm. Indeed, the expressionfrom (2) for x = −1 becomes(1− t)−B0(1− t)B1(1− t)−B2(1− t)B3 · · · = (1− t)−χ(M) =∞∑m=0(−1)mtm(−χ(M)m)=∞∑m=0tm(m + χ(M)− 1m),and hence the proof of Theorem 3 follows.Theorem 4. If M is a compact CW complex such that χ(M) = 0, then M (m)cannot be decomposed into cells of type Ci.Proof. If χ(M) = 0, then χ(M (m)) = 0. If M (m) can be decomposed into cells oftype Ci, then its Euler characteristic should be a positive number, which contradictsto χ(M (m)) = 0.Notice that if M is a torus, then it is a complex manifold, but its symmetricproduct cannot be decomposed into complex cells C0,C,C2, · · ·The Euler characteristic of the symmetric product 2993. Direct proof of the Theorem 3In this section we present a direct proof of Theorem 3, without using the Macdonald’sresults.Here Euler characteristic χ will denote the number of even dimensional cells oftype Ri minus the number of odd dimensional cells of the same type where all ofthe cells are disjoint.Proof. Notice that (5) is trivially satisfied for m = 1, because M (1) = M . So,without loss of generality, in the proof we assume that m ≥ 2. First, let us proveformula (5) for the cells C = Rn. The proof is by induction of n. If n = 0, i.e. Cis a point, then C(m) is also a point, and (5) is true because χ(C) = χ(C(m)) = 1.Further, if n = 1, i.e. C = R, then according to [17], C(m) is homeomorphic to thehalf space Hm and (5) holds because χ(C) = −1 and χ(C(m)) = 0. If C = R2, thenC(m) = R2m and (5) is true, because χ(C) = χ(C(m)) = 1.Let us assume that n ≥ 3. We should prove that χ((Rn)(m)) is equal to 0 if nis an odd number and it is 1 if n is an even number, for m > 1. In all calculationswe will neglect all sets whose Euler characteristic is 0 according to the inductiveassumption.Assume that n = 2k + 1 and the statement is true for 2k. We consider the spaceRn as a disjoint union of a hyperplane Σ and the corresponding two open half spacesΣ1 and Σ2. We prove that χ((R2k+1)(m)) = 0 by induction of m. Assume that itis true for 2, 3, · · · ,m− 1. According to the inductive assumption, it is sufficient toconsider only those cells where only one point, m points or no points are chosen inΣ1 (also Σ2). But if m points are chosen in Σ1 (resp. Σ2), then in Σ2 (resp. Σ1)there is no one chosen point. Hence there we should consider only these six cases:1) all points are chosen from Σ1, 2) all points are chosen from Σ2, 3) all points arechosen from Σ, 4) one point is chosen from Σ1, one point from Σ2 and the rest m−2points are chosen from Σ, 5) one point is chosen from Σ1 and the rest m− 1 pointsare chosen from Σ, 6) one point is chosen from Σ2 and the rest m − 1 points arechosen from Σ. Using the inductive assumption for n − 1 = dimΣ, for the Eulercharacteristic of (Rn)(m) we obtain the equalityχ((Rn)(m)) = χ(Σ(m)1 ) + χ(Σ(m)2 ) + χ(Σ(m)) + χ(Σ1)× χ(Σ2)× χ(Σ(m−2))+χ(Σ1)× χ(Σ(m−1)) + χ(Σ2)× χ(Σ(m−1))= 2χ((Rn)(m)) + 1 + 1− 1− 1and hence we obtain χ((Rn)(m)) = 0.Assume that n = 2k and the statement is true for 2k− 1. We consider the spaceRn as a disjoint union of a hyperplane Σ and two open half spaces Σ1 and Σ2 as inthe previous case. We prove that χ((R2k)(m)) = 1 by induction of m. Assume thatit is true for 2, 3, · · · ,m− 1. According to the inductive assumption, it is sufficientto consider only those cells where one point or no points are chosen in Σ. Hence we300 K.Trenčevskiobtain the following equalityχ((Rn)(m)) =m∑i=0χ(Σ(i)1 ) · χ(Σ(m−i)2 ) +m−1∑i=0χ(Σ(i)1 ) · χ(Σ(m−1−i)2 ) · χ(Σ)= 2χ((Rn)(m)) + (m− 1) + m(−1)2k−1 = 2χ((Rn)(m))− 1,and hence χ((Rn)(m)) = 1.To complete the proof it is sufficient to prove that if (5) is true for arbitrary mand disjoint sets X1 and X2 of cells of the CW complex M , then formula (5) is truefor X1 ∪X2. Let χ(X1) = p, χ(X2) = q and assume thatχ(X(i)1 ) =(i + p− 1i)for i = 0, 1, 2, · · · ,χ(X(i)2 ) =(i + q − 1i)for i = 0, 1, 2, · · · .Thenχ((X1 ∪X2)(m))= χ[X(m)1 ∪ (X(m−1)1 ×X2) ∪ (X(m−2)1 ×X(2)2 ) ∪ · · · ∪X(m)2]=m∑s=0χ(X(s)1 ) · χ(X(m−s)2 )=m∑s=0(s + p− 1s)(m− s + q − 1m− s)=m∑s=0(−1)s(−ps)(−1)m−s( −qm− s)= (−1)mm∑s=0(−ps)( −qm− s)= (−1)m(−p− qm)=(m + p + q − 1m).References[1] A.Dold, Homology of symmetric products and other functions of complexes, Ann.Math. 68(1958), 54–80.[2] S.Kera, On the permutation products of manifolds, Contribution to Algebra andGeometry 42(2001), 547–555.[3] S.Kera, One class of submanifolds of permutation products of complex manifolds,Math. Balkanica N.S. 15(2001), 275–284.The Euler characteristic of the symmetric product 301[4] S.D. Liao, On the topology of cyclic products on spheres, Tarns. Amer. Math. Soc.77(1954), 520–551.[5] I. G.Macdonald, The Poincaré polynomial of a symmetric product, Proc. Camb.Phil. Soc. 58(1962), 563–568.[6] R. 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The Euler characteristic of the symmetric product of a finite CW-complex

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